Check if a subset of matrices is a linear space

Question

Let $\mathbb{U}$ and $\mathbb{W}$ be subsets of $\mathbb{M}_2(\mathbb{R})$ - the linear space of all 2x2 matrices with real number entries.

$ \mathbb{U} = \{\begin{pmatrix} 3y-9x&x\\ y&z\\ \end{pmatrix} \mid x,y,z \in \mathbb{R}\}\\ \mathbb{W} = \{\begin{pmatrix} 6x & -3x \\ 8x & x \end{pmatrix} \mid x \in \mathbb{R}\}$

a) Prove that $\mathbb{U}$ and $\mathbb{W}$ are linear spaces with the usual matrix addition and scalar multiplication;

b) Find the basis of $\mathbb{U}$ and $\mathbb{W}$;

c) For which values of $\lambda, \mu \in \mathbb{R}$ is the matrix $A = (\begin{smallmatrix}\mu & -12 \\ 40 & \lambda\end{smallmatrix})$ an element of $\mathbb{U}$ and for which of $\mathbb{W}$?

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For a) I proved that $\mathbb{U}$ and $\mathbb{W}$ are subspaces of $\mathbb{M}_2(\mathbb{R})$, therefore they are linear spaces.

For b) I am not so sure what I did is correct so I'd like someone to confirm that for me. I did the following for the basis of $\mathbb{U}$: $(\begin{smallmatrix}3y-9x&x\\y&z\end{smallmatrix}) = (\begin{smallmatrix}3y-9x&0\\0&0\end{smallmatrix}) + (\begin{smallmatrix}0&x\\0&0\end{smallmatrix}) + (\begin{smallmatrix}0&0\\y&0\end{smallmatrix}) + (\begin{smallmatrix}0&0\\0&z\end{smallmatrix}) = (3y-9x)(\begin{smallmatrix}1&0\\0&0\end{smallmatrix}) + x(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}) + y(\begin{smallmatrix}0&0\\1&0\end{smallmatrix}) + z(\begin{smallmatrix}0&0\\0&1\end{smallmatrix}) = y(\begin{smallmatrix}3&0\\0&0\end{smallmatrix}) + x(\begin{smallmatrix}-9&0\\0&0\end{smallmatrix}) + x(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}) + y(\begin{smallmatrix}0&0\\1&0\end{smallmatrix}) + z(\begin{smallmatrix}0&0\\0&1\end{smallmatrix}) = x(\begin{smallmatrix}-9&1\\0&0\end{smallmatrix}) + y(\begin{smallmatrix}3&0\\1&0\end{smallmatrix}) + z(\begin{smallmatrix}0&0\\0&1\end{smallmatrix})$.

From this it follows that the basis is the set $\{(\begin{smallmatrix}-9&1\\0&0\end{smallmatrix}), (\begin{smallmatrix}3&0\\1&0\end{smallmatrix}), (\begin{smallmatrix}0&0\\0&1\end{smallmatrix})\}$.

Following a similar argument for $\mathbb{W}$ the basis is $\{(\begin{smallmatrix}6&-3\\8&1\end{smallmatrix})\}$.

c) For $A \in \mathbb{U}$ it means that $\mu = 3\cdot40 + 9\cdot12 = 228$ and $\lambda \in \mathbb{R}$ is unrestricted. For $A \in \mathbb{W}$ it means that $\mu = 6\lambda$, $40 = 8\lambda$, $-12 = -3\lambda$ which is impossible.

Answer 1

Blockquote From this it follows that the basis is the set $\{(\begin{smallmatrix}-9&1\\0&0\end{smallmatrix}), (\begin{smallmatrix}3&0\\1&0\end{smallmatrix}), (\begin{smallmatrix}0&0\\0&1\end{smallmatrix})\}$.

No, you have just shown that this is a generating set. Now you need to show that it is free (easily done bellow)

lets call those three matrices (which of course are vectors in $\mathbb{M}_2(\mathbb{R}))$ $u$, $v$, and $w$. We show that $(u, v, w)$ is free :

assume we have three reals $\alpha$, $\beta$ and $\gamma$ (say), such that $\alpha u + \beta v + \gamma w = 0$. That is to say :

$\left(\begin{matrix} -9 \alpha + 3 \beta & \alpha \\ \beta & \gamma\end{matrix}\right) = \left(\begin{matrix} 0 & 0 \\ 0 & 0\end{matrix}\right) $ This readily says : $\alpha = \beta = \gamma = 0$.

Hence, $(u, v, w)$ is free.